1. Chemistry 102(01) Spring 2008 Instructor: Dr. Upali Siriwardane
Office: CTH 311 Phone
Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;
                      Tu,Th,F  10: :00 a.m.: 
Test Dates:
March 19,  2008 (Test 1): Chapter 13
April 16, (Test 2): Chapters 14 & 15
May 12, 2008 Chapters & 18
Comprehensive Final Exam: May 14, 2008:Chapters 13, 14, 15, 16, 17 and 18

2. Chapter 17. Aditional Aqueous Equilibria
17.1 Buffer Solutions
17.2 Acid-Base Titrations
17.3 Acid Rain
17.4 Solubility Equilibria and the Solubility Product Constant, Ksp
17.5 Factors Affecting Solubility / Precipitation: Will It Occur?

3. CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
Hydrolysis
Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction.
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)
This type of reaction is given a special name.
Hydrolysis
The reaction of an anion with water to produce the conjugate acid and OH-.
The reaction of a cation with water to produce the conjugate base and H3O+.

4. How do you calculate pH of a salt solution?
Find out the pH, acidic or basic?
If acidic it should be a salt of weak base
If basic it should be a salt of weak acid
if acidic calculate Ka from Ka= Kw/Kb
if basic calculate Kb from Kb= Kw/Ka
Do a calculation similar to pH of a weak acid or base

5. What is the pH of 0. 5 M NH4Cl salt solution. (NH 3; Kb = 1
What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5) Find out the pH, acidic
if acidic calculate Ka from Ka= Kw/Kb
Ka= Kw/Kb = 1 x /1.8 x 10-5)
Ka= X 10-10
Do a calculation similar to pH of a weak acid

6. Continued NH4+ + H2O <==> H 3+O + NH3 [NH4+] [H3+O ] [NH3 ]
Ini. Con M M 0.00 M
Eq. Con x x x
[H 3+O ] [NH3 ]
Ka(NH4+) = =
[NH 4+] x2
; appro.: x
( x)

7. Continued x2 Ka(NH4+) = ----------- = 5.56 x 10 -10 0. 5
x2 = x x 0.5 = x
x= sqrt 2.78 x = x 10-5
[H+ ] = x = 1.66 x 10-5 M
pH = -log [H+ ] = - log x 10-5
pH = 4.77
pH of 0.5 M NH4Cl solution is 4.77 (acidic)

8. Common Ion Effect This is an example of Le Châtelier’s principle.
The shift in equilibrium caused by the addition of an ion formed from the solute.
Common ion
An ion that is produced by more than one solute in an equilibrium system.
Adding the salt of a weak acid to a solution of weak acid is an example of this.

9. Common Ion Effect Weak acid and salt solutions
E.g. HC2H3O2 and NaC2H3O2
Weak base and salt solutions
E.g. NH3 and NH4Cl.
H2O + C2H3O2- <==> OH- + HC2H3O (common ion)
H2O + NH <==> H3+O + NH3
(common ion)

10. HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Buffers
Solutions that resist pH change when small amounts of acid or base are added.
Two types
Mixture of weak acid and its salt
Mixture of weak base and its salt
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Add OH Add H3O+
shift to right shift to left
Based on the common ion effect.

11. Buffers
The pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two.
Henderson-Hasselbalch equation.
Easily derived from the Ka or Kb expression.
Starting with an acid
pH = pKa + log
Starting with a base
pH = 14 - ( pKb + log )
[A-]
[HA]
[HA]
[A-]

12. Henderson-Hasselbalch Equation
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[H3O+] [A-]
Ka =
[HA]
[H3O+] = Ka  ([HA]/[A-])
pH = pKa + log([A-]/[HA])
when the [A-] = [HA]
pH = pKa

13. Calcualtion of pH of Buffers Henderson Hesselbach Equation
[ACID]
pH = pKa - log
[BASE]
pH = pKa + log

14. Buffers and blood Control of blood pH
Oxygen is transported primarily by hemoglobin in the red blood cells.
CO2 is transported both in plasma and the red blood cells.
CO2 (aq) + H2O
H2CO3 (aq)
The bicarbonate
buffer is essential
for controlling
blood pH
H+(aq) + HCO3-(aq)

15. Buffer Capacity
Refers to the ability of the buffer to retard changes in pH when small amounts of acid or base are added
The ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity

16. Adding an Acid or Base to a Buffer

17. Buffer Systems

18. Titrations of Acids and Bases
Analyte
Titrant
analyte + titrant => products

19. Indicators
Acid-base indicators are highly colored weak acids or bases.
HIn In H+
color color 2
They may have more than one color transition.
Example. Thymol blue
Red Yellow Blue
One of the forms may be colorless - phenolphthalein (colorless to pink)

20. Acid-Base Indicator Weak acid that changes color with changes in pH
HIn + H2O  H3O+ + In-
acid base
color color
[H3O+][In-]
Ka =
[HIn]
They may have more than one color transition.
Example. Thymol blue
Red Yellow Blue

21. What is an Indicator?
Indicator is an weak acid with different Ka, colors to the acid and its conjugate base. E.g. phenolphthalein
HIn <===> H In-
colorless pink
Acidic colorless
Basic pink

22. Selection of an indicator for a titration
a) strong acid/strong base
b) weak acid/strong base
c) strong acid/weak base
d) weak acid/weak base
Calculate the pH of the solution at he equivalence point or end point

23. pH and Color of Indicators

24. Red Cabbage as Indicator

25. Indicator examples
Acid-base indicators are weak acids that undergo a color change at a known pH. pH
phenolphthalein

26. Titration Apparatus
Buret delivering base to a flask containing an acid. The pink color in the flask is due to the phenolphthalein indicator.

27. Endpoint vs. Equivalence Point
point where there is a physical change, such as color change, with the indicator
Equivalence Point
# moles titrant = # moles analyte
#molestitrant=(V  M)titrant
#molesanalyte=(V  M)analyte

28. Acid-Base Indicator Behavior
acid color shows when
[In-]
[HIn]
[H3O+][In-] 1
=  [H3O+] = Ka
[HIn] 10
base color shows when
[H3O+][In-]
= 10  [H3O+] = Ka
[HIn]
[In-]
[HIn]

29. Indicator pH Range acid color shows when pH + 1 = pKa
and base color shows when
pH = pKa
Color change range is
pKa = pH  or pH = pKa  1

30. Titration curves Acid-base titration curve
A plot of the pH against the amount of acid or base added during a titration.
Plots of this type are useful for visualizing a titration.
It also can be used to show where an indicator undergoes its color change.

31. Indicator and Titration Curve 0.1000 M HCl vs 0.1000 M NaOH

32. EXAMPLE: Derive the titration curve for the titration of mL of M HCl with 0.00, 15.00, 35.00, and mL of M NaOH.
at 0.00 mL of NaOH added, initial point
[H3O+] = CHCl = M
pH =

33. ((35.00mL)  (0.1000M)) - ((15.00mL)  (0.1000M))
EXAMPLE: Derive the titration curve for the titration of mL of M HCl with 0.00, 15.00, 35.00, and mL of M NaOH.
at mL of NaOH added
Va  Ma > Vb  Mb thus
(Va  Ma) - (Vb  Mb)
[H3O+] =
(Va + Vb)
((35.00mL)  (0.1000M)) - ((15.00mL)  (0.1000M)) =
( )mL
=  10-2 M
pH =

34. at equivalence point of a strong acid - strong base titration
EXAMPLE: Derive the titration curve for the titration of mL of M HCl with 0.00, 15.00, 35.00, and mL of M NaOH.
at mL of NaOH added
Va  Ma = Vb  Mb , equivalence point
at equivalence point of a
strong acid - strong base titration
pH º

35. EXAMPLE: Derive the titration curve for the titration of mL of M HCl with 0.00, 15.00, 35.00, and mL of M NaOH.
at mL of NaOH added
Vb  Mb > Va  Ma , post equvalence point
(Vb  Mb) - (Va  Ma)
[OH-] =
(Va + Vb)
((50.00mL)  (0.1000M)) - ((35.00mL)  (0.1000M)) =
( )mL
=  10-2 M
pOH =
pH = = 12.25

36. Titration of Strong Acid with Strong Base
equivalence point x

37. Titration of Weak Acid with Strong Base

38. Effect of Acid Strength on Titration Curve

39. Titration of Weak Base with Strong Acid

40. Titration of Diprotic Weak Acid with Strong Base

41. pH range of Indicators litmus (5.0-8.0) bromothymole blue (6.0-7.6)
methyl red ( )
thymol blue ( )
phenolphthalein ( )
thymolphthalein ( )

42. Acid Rain acid rain is defined as rain with a pH < 5.6
pH = 5.6 for rain in equilibrium with atmospheric carbon dioxide

43. Sulfuric Acid from Sulfur burning
SO2
S + O2 => SO2
SO3
2 SO2 + O2 => 2 SO3
Sulfuric Acid
SO3 + H2O => H2SO4

44. 2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq)
Nitric Acid
2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq)

45. How Acid Precipitation Forms

46. Acid Precipitation in U.S.

47. Solubility Product
solubility-product - the product of the solubilities
solubility-product constant => Ksp
constant that is equal to the solubilities of the ions produced when a substance dissolves

48. Solubility Product Constant
In General:
AxBy  xA+y + yB-x
[A+y]x [B-x]y
K =
[AxBy]
[AxBy] K = Ksp = [A+y]x [B-x]y
Ksp = [A+y]x [B-x]y
For silver sulfate
Ag2SO4  2 Ag+ + SO4-2
Ksp = [Ag+]2[SO4-2]

49. Solubility Product Constant Values

50. Dissolving Slightly Soluble Salts Using Acids
Insoluble salts containing anions of Bronsted-Lowry bases can be dissolved in solutions of low pH

51. Calcium Carbonate Dissolved in Acid
Limestone Dissolving in Ground Water
CaCO3(S) + H2O + CO2 => Ca+2(aq) + 2 HCO3-(aq)
Stalactite and Stalagmite Formation
Ca+2(aq) + 2 HCO3-(aq) => CaCO3(S) + H2O + CO2

52. The Common Ion Effect common ion
second source which is completely dissociated
In the presence of a second source of the ion, there will be less dissolved than in its absence
common ion effect
a salt will be less soluble if one of its constitutent ions is already present in the solution

53. (x)(x) = Ksp = [Ag+][Cl-] = 1.82  10-10M2
EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?
AgCl  Ag+ + Cl-
Ksp = [Ag+][Cl-] = 1.82  10-10M2
let x = molar solubility = [Ag+] = [Cl-]
(x)(x) = Ksp = [Ag+][Cl-] = 1.82  10-10M2
x = 1.35  10-5M

54. EXAMPLE: What is the molar solubility of AgCl in pure water and in 1
EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl?
AgCl  Ag+ + Cl-
Ksp = [Ag+][Cl-] = 1.82  10-10M2
let x = molar solubility = [Ag+]
[Cl-] = 1.0 M
Ksp = [Ag+][Cl-] = (x)(1.0M) = 1.82  10-10M2
x = 1.82  10-10M

55. Formation of Complexes
ligand - Lewis base
complexes - product of Lewis acid-base reaction
Ag+(aq) NH3(aq)  [Ag(NH3)2(aq)]+
Ag+(aq) + Cl-(aq)  AgCl(s)
AgCl(s) NH3(aq)  [Ag(NH3)2(aq)]+ + Cl-(aq)

56. Sodium Thiosulfate Dissolves Silver Bromide

57. Formation Constants

58. Amphoterism

59. Reactant Quotient, Q Reactant Quotient, Q
ion product of the initial concentration
same form as solubility product constant
Q < Ksp - no precipitate forms an unsaturate
solution
Q > Ksp - precipitate may form to restore condition of saturated solution
Q = Ksp - no precipitate forms, saturated

60. Will Precipitation Occur?

61. magnesium ammonium phsphate
Kidney Stones
Kidney stones are normally composed of:
calcium oxalate
calcium phosphate
magnesium ammonium phsphate

62. Calcium Oxalate
Precipitate formed from calcium ions from food rich in calcium, dairy products, and oxalate ions from fruits and vegetables
Ca C2O4-2  CaC2O4

63. PRECIPITATION REACTIONS

64. Analysis of Silver Group
These salts formed are insoluble, they do dissolve to some SLIGHT extent.
AgCl(s) Ag+(aq) + Cl-(aq)
When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

65. Analysis of Silver Group
AgCl(s) Ag+(aq) + Cl-(aq)
When solution is SATURATED, expt. shows that [Ag+] = 1.67 x 10-5 M.
(SOLUBILITY) of AgCl.
What is [Cl-]?

66. Analysis of Silver Group
AgCl(s) Ag+(aq) + Cl-(aq)
Saturated solution has [Ag+] = [Cl-] = 1.67 x 10-5 M
Use this to calculate Ksp
Ksp = [Ag+] [Cl-]
= (1.67 x 10-5)(1.67 x 10-5)
= x 10-10

67. Ksp = solubility product
Because this is the product of “solubilities”, we call it
Ksp = solubility product constant
See Table in the Text

68. Lead(II) Chloride
PbCl2(s) Pb2+(aq) Cl-(aq)
Ksp = 1.9 x 10-5

69. Solubility of Lead(II) Iodide
Consider PbI2 dissolving in water
PbI2(s) Pb2+(aq) I-(aq)
Calculate Ksp if solubility = M
Solution
1. Solubility = [Pb2+] = 1.30 x 10-3 M
[I-] = 2 x [Pb2+] = 2.60 x 10-3 M
2. Ksp = [Pb2+] [I-]2
= [Pb2+] {2 • [Pb2+]}2
= 4 [Pb2+]3 = 4 (solubility)3
Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9

70. Precipitating an Insoluble Salt
Hg2Cl2(s) Hg22+(aq) Cl-(aq)
Ksp = 1.1 x = [Hg22+] [Cl-]2
If [Hg22+] = M, what [Cl-] is req’d to just begin the precipitation of Hg2Cl2?
(maximum [Cl-] in M Hg22+ without forming Hg2Cl2?)

71. Precipitating an Insoluble Salt
Hg2Cl2(s) Hg22+(aq) Cl-(aq)
Ksp = 1.1 x = [Hg22+] [Cl-]2
Recognize that
Ksp = product of maximum ion concs.
Precip. begins when product of ion concs. EXCEEDS the Ksp.

72. Precipitating an Insoluble Salt
Hg2Cl2(s) Hg22+(aq) Cl-(aq)
Ksp = 1.1 x = [Hg22+] [Cl-]2
Solution
[Cl-] that can exist when [Hg22+] = M,
If this conc. of Cl- is just exceeded, Hg2Cl2 begins to precipitate.

73. Precipitating an Insoluble Salt
Hg2Cl2(s) Hg22+(aq) Cl-(aq)
Ksp = 1.1 x 10-18
Now raise [Cl-] to 1.0 M. What is the value of [Hg22+] at this point?
Solution
[Hg22+] = Ksp / [Cl-]2
= Ksp / (1.0)2 = 1.1 x M
The concentration of Hg22+ has been reduced by 1016 !

74. Separating Metal Ions Cu2+, Ag+, Pb2+
Cl-
Insoluble PbCl2 AgCl
Soluble CuCl2
Heat
Soluble PbCl2
Insoluble AgCl
Ksp Values
AgCl x 10-10
PbCl x 10-5
PbCrO4 1.8 x 10-14
CrO4-2
Insoluble PbCrO4

75. Separating Salts by Differences in Ksp
A solution contains M Ag+ and Pb2+. Add CrO42- to precipitate Ag2CrO4 (red) and PbCrO4 (yellow). Which precipitates first?
Ksp for Ag2CrO4 = 9.0 x 10-12
Ksp for PbCrO4 = 1.8 x 10-14
Solution
The substance whose Ksp is first exceeded precipitates first.
The ion requiring the smaller amount of CrO42- ppts. first.

76. Separating Salts by Differences in Ksp
Solution
Calculate [CrO42-] required by each ion.
[CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+]
= 1.8 x / = 9.0 x M
[CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2
= 9.0 x / (0.020)2 = 2.3 x 10-8 M
PbCrO4 precipitates first.

77. Separating Salts by Differences in Ksp
How much Pb2+ remains in solution when Ag+ begins to precipitate?
Solution
We know that [CrO42-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4.
What is the Pb2+ conc. at this point?
[Pb2+] = Ksp / [CrO42-] = 1.8 x / 2.3 x 10-8 M
= 7.8 x 10-7 M
Lead ion has dropped from M to < 10-6 M

78. Common Ion Effect Adding an Ion “Common” to an Equilibrium
PbCl2(s) Pb2+(aq) Cl-(aq)
NaCl Na+(aq) + Cl- (aq)

79. The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure water and (b) in M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) Ba2+(aq) SO42-(aq)
Solution
a) Solubility in pure water = [Ba2+] = [SO42-] = x
Ksp = [Ba2+] [SO42-] = x2
x = (Ksp)1/2 = 1.1 x 10-5 M

80. The Common Ion Effect BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = 1.1 x 10-10
Solution
b) Now dissolve BaSO4 in water already containing M Ba2+.
Which way will the “common ion” shift the equilibrium? ___
Will solubility of BaSO4 be less than or greater than in pure water?___

81. The Common Ion Effect BaSO4(s) Ba2+(aq) + SO42-(aq) Solution
initial
change
equilib.

82. The Common Ion Effect
Calculate the solubility of BaSO4 in (a) pure water and (b) in M Ba(NO3)2.
Ksp for BaSO4 = 1.1 x 10-10
BaSO4(s) Ba2+(aq) SO42-(aq)
Solution
[Ba2+] [SO42-]
initial
change + y y
equilib y y

83. The Common Ion Effect Ksp = [Ba2+] [SO42-] = (0.010 + y) (y)
Because y << x (1.1 x 10-5 M) y  Therefore,
Ksp = 1.1 x = (0.010)(y)
y = 1.1 x 10-8 M = solubility in presence of added Ba2+ ion.
Le Chatelier’s Principle is followed!